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| 183_notes:examples:walking_in_a_boat [2014/10/01 05:39] – pwirving | 183_notes:examples:walking_in_a_boat [2014/10/01 05:51] (current) – pwirving | ||
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| →rcm=1Mtot(∑imi→ri) in 1D, | →rcm=1Mtot(∑imi→ri) in 1D, | ||
| - | Therefore xcm,i initial | + | Therefore xcm,i initial |
| xcm,i=M(D+L2)+m(D+L)M+m | xcm,i=M(D+L2)+m(D+L)M+m | ||
| - | In the final state, we don't know x, but we know that xcm,f=xcm,i So we'll just use the unknown x. | + | In the final state, we don't know x, but we know that xcm,f=xcm,i So we'll just use the unknown x and insert the distance into the distances for boat and person relative to the origin. |
| xcm,f=M(x+D+L2)+m(x+D)M+m | xcm,f=M(x+D+L2)+m(x+D)M+m | ||
| + | |||
| + | As indicated the center of the mass of the system as not changed position therefore: | ||
| xcm,f=xcm,i | xcm,f=xcm,i | ||
| + | |||
| + | So we can relate the equations for initial and final to each other: | ||
| M(x+D+L2)+m(x+D)M+m=M(D+L2)+m(D+L)M+m | M(x+D+L2)+m(x+D)M+m=M(D+L2)+m(D+L)M+m | ||
| - | Same denominator | + | Both of these equations have the same denominator |
| M(x+D+L2)+m(x+D)=M(D+L2)+m(D+L) | M(x+D+L2)+m(x+D)=M(D+L2)+m(D+L) | ||
| - | Solve for x, | + | Multiply out to solve for x: |
| - | $M_x + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ | + | $Mx + M(D + \dfrac{L}{2}) + mx + mD = M(D+\dfrac{L}{2}) + mD + mL$ |
| - | Cancel like terms | + | Cancel like terms we get: |
| Mx+mx=mL | Mx+mx=mL | ||
| Line 102: | Line 106: | ||
| (M+m)x=mL | (M+m)x=mL | ||
| + | |||
| + | Therefore: | ||
| x=(mM+m)L | x=(mM+m)L | ||
| Line 108: | Line 114: | ||
| Does this make sense? | Does this make sense? | ||
| + | |||
| + | If you want to check whether something makes sense a good start is to check the units: | ||
| Units (x)=m (mM+m) = unitless | Units (x)=m (mM+m) = unitless | ||
| - | If M is really big then x≡0, think oil thanker | + | Therefore m is the remaining unit. |
| + | |||
| + | Another check is that we know that if M is really big then x≡0, think of a similar scenario to one just discussed occurring on a oil thanker. | ||
| x=(mM+m)L≡mML≡0 when M>>m | x=(mM+m)L≡mML≡0 when M>>m | ||
| - | If M = 0 then x ≡L, no boat limit | + | Another check would be to check what happens when M = 0 then x ≡L, i.e. if there was no boat. |
| x=(mM+m)L≡mML≡L when m>>M | x=(mM+m)L≡mML≡L when m>>M | ||
| So the motion of the center of mass of a system is dictated by the net external force. | So the motion of the center of mass of a system is dictated by the net external force. | ||